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In a class of 100 students, the mean score of the exam is 75, with a standard deviation of 6. What is the maximum number of students who could have made 100 on the exams?

What I tried

$P(X = 100) = P(X>99)\leq (1/99)E(X) = 75$ students. Apparently this is not correct.

Thanks,

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I'll assume that the possible scores are integers $0$ to $100$ (fractional scores are not allowed). Suppose $x_i$ students get score $i$, $i = 0 \ldots 100$. We have $$ \eqalign{\sum_{i=0}^{100} x_i &= 100\cr \sum_{i=0}^{100} i x_i &= 7500\cr \sum_{i=0}^{100} i^2 x_i &= 100 (6^2 + 75^2) = 566100\cr \text{all}\ x_i \ge 0} $$ and you want to maximize $x_{100}$.

As a linear programming problem, this has optimal solution $x_{73} = 1100/27 \approx 40.74$, $x_{74} = 700/13 \approx 53.85$, $x_{100} = 1900/351 \approx 5.41$. Of course we want integer solutions, but this does tell us it's impossible to do better than $x_{100} = 5$. It turns out there are integer solutions with $x_{100} = 5$, e.g.:

$$ \eqalign{x_{67} &= 3\cr x_{71} &= 2\cr x_{72} &= 1\cr x_{73} &= 17\cr x_{74} &= 69\cr x_{76} &= 1\cr x_{77} &= 1\cr x_{85} &= 1\cr x_{100} &= 5\cr \text{all other}\ x_i &= 0\cr} $$ (I found this using Cplex)

Thus the maximum possible number of students getting $100$ is $5$.

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